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Chapter 9: Problem 32

You are to design a rotating cylindrical axle to lift \(800-\mathrm{N}\) buckets of cement from the ground to a rooftop 78.0 \(\mathrm{m}\) above the ground. The buckets will be attached to a hook on the free end of a cable that wraps around the rim of the axle; as the axle turms, the buckets will rise. (a) What should the diameter of the axle be in order to raise the buckets at a steady 2.00 \(\mathrm{cm} / \mathrm{s}\) when it is turning at 7.5 \(\mathrm{rpm} ?\) (b) If instead the axle must give the buckets an upward acceleration of \(0.400 \mathrm{m} / \mathrm{s}^{2},\) what should the angular acceleration of the axle be?

Short Answer

Expert verified
(a) Diameter 鈮� 5.1 cm; (b) Angular acceleration 鈮� 15.7 rad/s虏.

Step by step solution

01

Understanding the Problem

We need to determine two things: (a) the diameter of the rotating axle that will allow a bucket to rise at a steady speed of 2.00 cm/s when turning at 7.5 rpm, and (b) the angular acceleration of the axle when the bucket should rise with an upward acceleration of 0.400 m/s虏.
02

Convert Units

For part (a), convert the linear speed from cm/s to m/s: \(v = 2.00 \text{ cm/s} = 0.020 \text{ m/s}\). Also convert the rotational speed from rpm to radians per second: \(7.5 \text{ rpm} = \frac{7.5 \times 2\pi}{60} \text{ rad/s} = 0.7854 \text{ rad/s}\).
03

Diameter from Steady Rise (Part a)

The relationship between linear speed \(v\) and angular speed \(\omega\) is \(v = r \omega\), where \(r\) is the radius of the axle. Solving for \(r\), we have \(r = \frac{v}{\omega} = \frac{0.020}{0.7854} = 0.0255 \text{ m}\). Therefore, the diameter \(D\) is \(2r = 0.0510 \text{ m}\) or 5.10 cm.
04

Determine Angular Acceleration (Part b)

For part (b), relate the linear acceleration \(a = 0.400 \text{ m/s}^2\) to angular acceleration \(\alpha\) using \(a = r \alpha\). Using \(r = 0.0255 \text{ m}\) from step 3, \(\alpha = \frac{a}{r} = \frac{0.400}{0.0255} \approx 15.69 \text{ rad/s}^2\).

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Rotational Motion
Rotational motion describes the motion of objects that spin or rotate about an axis. In many real-world applications like engines, wheels, and our exercise's rotating cylindrical axle, rotational motion is central to their function. Key aspects include:

- **Rotational Speed**: This is how fast an object turns, often measured in revolutions per minute (rpm) or radians per second (rad/s). - **Radius and Diameter**: These define the size of the rotating object. The diameter is twice the radius, and in our problem, finding the correct radius is essential to ensure the axle lifts the cement buckets correctly.
Understanding these concepts helps predict how the axle will behave as it turns. It ensures that the rotational speed aligns with the linear requirements of the task鈥攔aising a bucket at a specified velocity. Grasping these basics can provide a deeper insight into how rotations facilitate mechanical lifting tasks.
Angular Acceleration
Angular acceleration refers to how quickly the rotational speed of an object changes. In simpler terms, it's similar to the concept of acceleration in linear motion, but it applies to rotation. In our axle lifting problem, angular acceleration comes into play when we want to lift the bucket not just at a constant speed but with an upward acceleration. The formula connecting linear acceleration (\( a \)) and angular acceleration (\( \alpha \)) is: \[ a = r \alpha \] where \( r \) is the radius.

This relationship ensures mechanical systems coordinate their linear and rotational parts effectively. An increased angular acceleration means the axle spins faster over time, increasing the upward acceleration of the bucket. The calculated angular acceleration of approximately 15.69 rad/s虏 ensures the bucket accelerates upwards as required. This concept is crucial in many engineering and design applications across machines and vehicles.
Unit Conversion
Unit conversion is essential for solving physics problems accurately, as it ensures all measurements are in compatible units. In our exercise, converting units was necessary to reach the solution.

- **Linear Speed Conversion**: From cm/s to m/s by dividing the speed in cm/s by 100 (e.g., 2.00 cm/s = 0.020 m/s). - **Rotational Speed Conversion**: From rpm to rad/s involves using the formula:\[ \text{rad/s} = \frac{\text{rpm} \times 2\pi}{60} \] This accounts for converting revolutions (a full circle of \( 2\pi \) radians) to per minute, then to per second.
Accurate unit conversion prevents errors that could lead to incorrect results. Units must be consistent across equations so that calculations, such as finding radius or angular acceleration, yield the expected results. Mastering unit conversion is a valuable skill for tackling a wide range of physics problems.

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Most popular questions from this chapter

Engineers are designing a system by which a falling mass \(m\) imparts kinetic energy to a rotating uniform drum to which it is attached by thin, very light wire wrapped around the rim of the drum (Fig. 9.34\()\) . There is no appreciable friction in the axle of the drum, and everything starts from rest. This system is being tested on earth, but it is to be used on Mars, where the acceleration due to gravity is 3.71 \(\mathrm{m} / \mathrm{s}^{2} .\) In the earth tests, when \(m\) is set to 15.0 \(\mathrm{kg}\) and allowed to fall through \(5.00 \mathrm{m},\) it gives 250.0 \(\mathrm{J}\) of kinetic energy to the drum. (a) If the system is operated on Mars, through what distance would the \(15.0-0\) mass have to fall to give the same amount of kinetic energy to the drum? (b) How fast would the 15.0 \(\mathrm{kg}\) mass be moving on Mars just as the drum gained 250.0 \(\mathrm{J}\) of kinetic energy?

A flywheel with a radius of 0.300 \(\mathrm{m}\) starts from rest and accelerates with a constant angular acceleration of 0.600 \(\mathrm{rad} / \mathrm{s}^{2}\) . Compute the magnitude of the tangential acceleration, the radial acceleration, and the resultant acceleration of a point on its rim (a) at the start; (b) after it has turned through \(60.0^{\circ} ;\) (c) after it has turned through \(120.0^{\circ} .\)

A straight piece of reflecting tape extends from the center of a wheel to its rim. You darken the room and use a camera and strobe unit that flashes once every 0.050 s to take pictures of the wheel as it rotates counterclockwise. You trigger the strobe so that the first flash \((t=0)\) occurs when the tape is horizontal to the right at an angular displacement of zero. For the following situations draw a sketch of the photo you will get for the time exposure over five flashes (at \(t=0,0.050 \mathrm{s}, 0.100 \mathrm{s}, 0.150 \mathrm{s},\) and 0.200 \(\mathrm{s} )\) and graph \(\theta\) versus \(t\) and \(\omega\) versus \(t\) for \(t=0\) to \(t=0.200 \mathrm{s}\) (a) The angular velocity is constant at 10.0 rev \(/ \mathrm{s}\) . (b) The wheel starts from rest with a constant angular acceleration of 25.0 rev \(/ \mathrm{s}^{2}\) . (c) The wheel is rotating at 10.0 rev \(/ \mathrm{s}\) at \(t=0\) and changes angular velocity at a constant rate of \(-50.0 \mathrm{rev} / \mathrm{s}^{2}\) .

A cylinder with radius \(R\) and mass \(M\) has density that increases linearly with distance \(r\) from the cylinder axis, \(\rho=\alpha r\) where \(\alpha\) is a positive constant. (a) Calculate the moment of inertia of the cylinder about a longitudinal axis through its center in terms of \(M\) and \(R .\) (b) Is your answer greater or smaller than the moment of inertia of a cylinder of the same mass and radius but of uniform density? Explain why this result makes qualitative sense.

Trip to Mars. You are working on a project with NASA to launch a rocket to Mars, with the rocket blasting off from earth when earth and Mars are aligned along a straight line from the sun. If Mars is now \(60^{\circ}\) shead of earth in its orbit around the sun, when should you launch the rocket? (Note: All the planets orbit the sun in the same direction, 1 year on Mars is 1.9 earth- years, and assume circular orbits for both planets.)
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